Discover

Share

Buzz Blog

Olympic Women Ski Jump Equally Far on the Moon

Tuesday, February 11, 2014

Right now, for the first time in Olympic history, women are taking to the ski jumping slopes to compete for Olympic gold. If they were on the Moon how far might they fly?

Would they surpass the Moon’s escape velocity and go soaring through space? Or would they manage to circle around the circumference of the Moon before landing? Neither. The provocative, counterintuitive answer is that they would travel about the same distance!

The Holmenkillen ski Jump in Oslo, Norway. Credit: Mathias Stang

The women are competing on the smaller of the two hills, the normal hill, at the RusSki Gorki Jumping Center. The center’s normal hill is a K95. The K stands for K point, or critical point, which is a target for skiers, and the 95 means that the target point is 95 meters away from the launch point.

If skiers jump a distance of exactly 95 meters (about 312 feet) then they earn 60 points. They are either awarded or penalized points for how many meters they surpass or fall short of the K point.

Ski jumpers have a number of tactics they use to make sure they fly longer and farther than any of their competitors. Ski length, body position, weight, strength and air resistance are some of the factors that determine how fast the skier flies off the ramp, how long they’re in the air and how far they travel.

It is very difficult to calculate exactly where a skier will touch ground on Earth because of air resistance and the fact that it both helps and hurts the athletes. From the second they start down the ramp, air resistance is acting against them. But once the slope drops out from under them, they can generate lift while in the air, which is why they make a V-shape with their skis. It’s hard to know exactly how much the drag verses the lift affects a skier’s distance for a given jump since weather conditions affect wind speed and direction.

On the Moon, however, the skiers need not worry about air resistance. Since the Moon has a lower gravity, you might think that the skiers would travel farther because they’re in the air for a longer period of time. But my calculations show that if the skiers were to jump from the same slope on the Moon, they would be moving slower off the jump and the ratio of the distance they travel on the Moon verses on Earth is one, meaning they travel the same distance!

To calculate how fast the skiers are traveling when they leave the ramp, I use simple energy conservation.

On Earth

On Moon

Ratio of the two gives me:

This is the ratio for both the velocity in the x and y direction. We ultimately want to know what is the ratio for the horizontal distance traveled, which I represent as dxE/dxM.

To calculate how far the skier travels along the x-direction I use the simple relation:

Where dx is the distance traveled, v is the velocity in the x direction (which remains constant in the absence of air resistance) and dt is the time in the air.

The ratio of this relation on Earth and the Moon is:

I already calculated the ratio of the velocities. All I need now is the ratio of the time the skiers are in the air on Earth verses the Moon. I can calculate this ratio using the relation that

Where dv is the change in velocity in the y direction, a is acceleration due to gravity and dt is the time in the air.

After solving for dt, the ratio of the time the skier is in the air on the Earth and Moon is:

Now subbing this into my equation for the ratio of distance traveled, I get:

We calculated earlier that :

and the ratio of the velocity in the y-direction at any given point in time is equal to this same ratio, so:

The ratio between the distance traveled on Earth versus the Moon is equal to one, which means that the distance ski jumpers travel on the Moon is the same for the distance they travel on Earth. Basically, the ski jumpers would perform the same, just in slow motion.

In reality the horizontal and vertical velocity on Earth will depend on air resistance. So, the distance ski jumpers travel will either be slightly shorter or longer than on the Moon depending on how kind the wind is that day.

6 Comments:

Anonymous said...

This blog relates to what I'm currently learning in my physics class. I feel that as long as the ratios are correct this makes perfect sense. Air pressure and velocity would need to have the same impact on the skier in order for the distance to be the same. We see on earth air resistance would effect the skier in both a positive way. While air resistance would not be a factor on the moon, having a slower velocity if equal to the effect of air resistance on earth could cause the skier to land in exactly the same distance on both the moon and the earth. - Jake Lawrence

Wednesday, February 12, 2014 at 7:42 PM

Anonymous said...

Retracting previous comments. The solution here is fine, as I'm sure the author already knows... :) The ratios dvy_e/dvy_m and v_e/v_m definitely are equal though they are solved for differently.

Wednesday, February 12, 2014 at 6:49 PM

Anonymous said...

I think the error in the solution presented is in the ratio used for dv_e/dv_m. It assumes the same energy conservation relations used to solve for v_e/v_m, which is incorrect because one now needs to include kinetic energy for the initial energy after the ramp. I would keep everything else in the solution the same and replace with the relation dv_e/dv_m >= sqrt(g_e/g_m). (It comes from energy conservation, with initial kinetic energy, and the triangle inequality (dv)^2 >= v_f^2 - v_0^2 = 2g_eh.) The '=' in this relation is for the case that v_e,v_m=0 and leads to them flying the same distance. All other cases (v_e, v_m not equal 0, which is what we have) lead to traveling a greater distance on earth!

Wednesday, February 12, 2014 at 5:49 PM

Jonathan Leon said...

This comment has been removed by the author.

Wednesday, February 12, 2014 at 5:33 PM

Anonymous said...

This is great idea to write about, but there's one error in your ideal case solution (i.e. no air resistance) that is a little disappointing. Everything is ok until you start to solve for the time spent in the air after the ramp. Solve for the ratio of these time explicitly, assuming the distance fallen after jump is the same on earth and the moon and using the velocities already solve for, and you'll get a very different answer

Wednesday, February 12, 2014 at 4:50 PM

Anonymous said...

This blog post is not only relevant to current events but to the introduction to physics class I am currently in. It is discussed here that velocity, time, distance, and gravity are important to get to the 95 meter mark, or further. Air resistance and gravity are important for this problem if it occurs on Earth. The air resistance could create a force against the skier and slow the velocity at which she is falling. Some have experienced this themselves when skiing and going off a jump. Since this problem is calculated on the moon where air resistance and gravity are not a factor it becomes easier to calculate. We learned that air resistance is not always a significant enough factor to calculate in, but in this case it happens to be necessary for a more accurate answer.